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A function $\phi : G \mapsto H$ is said to be an homomorphism iff

$\forall a,b \in G$, $\phi(ab) = \phi(a)\phi(b)$

Is this a complete definition of a group homomorphism. Do I not need to state one more condition that $\phi(e) =e$, where $e$ is the identity element of groups $G$ and $H$.

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  • $\begingroup$ $\phi(e_G)=e_H$ is a consequence of this definition. $\endgroup$ – Lord Shark the Unknown Mar 24 '18 at 8:26
  • $\begingroup$ @ Lord Shark the Unknown did you mean $\phi(e) = \phi(g.g^{-1}) =\phi(g) \phi(g)^{-1} = e$ $\endgroup$ – user437890 Mar 24 '18 at 8:30
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You have that $$e = \phi(e) \phi(e)^{-1} = \phi(e e) \phi(e)^{-1}= \phi(e) \phi(e) \phi(e)^{-1} = \phi(e).$$ Therefore one doesn't need that to be part of the axioms. Hope that helped you :)

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